在Hive SQL中计算最大连续活跃天数,可按以下思路和步骤进行:
### 思路分析
要计算最大连续活跃天数,关键在于将连续活跃的日期分组,然后统计每组的天数,最后找出最大的连续天数。可使用开窗函数`row_number`结合日期运算来实现分组。
### 示例代码
以引用[3]的数据为例,先创建测试表并插入数据:
```sql
-- 创建测试表
create table tmpdb.test_01 as
select '1001' as user_id, '2017-01-01' as login_date
union all
select '1001' as user_id, '2017-01-02' as login_date
union all
select '1001' as user_id, '2017-01-04' as login_date
union all
select '1001' as user_id, '2017-01-06' as login_date
union all
select '1001' as user_id, '2017-01-07' as login_date;
```
接着编写SQL计算最大连续活跃天数:
```sql
-- 计算最大连续活跃天数
SELECT
user_id,
MAX(consecutive_days) as max_consecutive_days
FROM (
SELECT
user_id,
grp,
COUNT(*) as consecutive_days
FROM (
SELECT
user_id,
login_date,
-- 将日期减去对应的排名,连续日期相减后结果相同,用于分组
date_sub(login_date, row_number() over (partition by user_id order by login_date)) as grp
FROM
tmpdb.test_01
) t1
GROUP BY
user_id, grp
) t2
GROUP BY
user_id;
```
### 代码解释
1. **内层子查询**:使用`row_number`函数对每个用户的登录日期进行排名,再用`date_sub`函数将登录日期减去排名,得到一个新的字段`grp`。连续活跃的日期相减后`grp`值相同,这样就把连续活跃的日期分到了同一组。
2. **中间子查询**:按用户和`grp`分组,统计每组的记录数,即连续活跃的天数。
3. **外层查询**:按用户分组,找出每个用户的最大连续活跃天数。
### 另一种示例
以引用[5]的数据为例,也可按上述思路计算最大连续活跃天数:
```sql
-- 先创建表并插入数据
CREATE TABLE user_activity (
user_id INT,
activity_date STRING
);
INSERT INTO user_activity VALUES
(1, '2022-07-11'),
(1, '2022-07-12'),
(1, '2022-07-13'),
(1, '2022-07-14'),
(1, '2022-07-15'),
(1, '2022-07-16'),
(1, '2022-07-17'),
(1, '2022-07-18'),
(2, '2022-07-12'),
(2, '2022-07-13'),
(2, '2022-07-14'),
(2, '2022-07-15'),
(2, '2022-07-16'),
(2, '2022-07-21'),
(2, '2022-07-22'),
(3, '2022-08-01'),
(3, '2022-08-02'),
(3, '2022-08-03'),
(3, '2022-08-04'),
(3, '2022-08-05'),
(3, '2022-08-06'),
(3, '2022-08-07'),
(3, '2022-08-08'),
(3, '2022-08-09'),
(3, '2022-08-10'),
(3, '2022-08-11');
-- 计算最大连续活跃天数
SELECT
user_id,
MAX(consecutive_days) as max_consecutive_days
FROM (
SELECT
user_id,
grp,
COUNT(*) as consecutive_days
FROM (
SELECT
user_id,
activity_date,
date_sub(activity_date, row_number() over (partition by user_id order by activity_date)) as grp
FROM
user_activity
) t1
GROUP BY
user_id, grp
) t2
GROUP BY
user_id;
```